x^n*x^n-3x^n+1*x^n-1

1.求幂级数 x/(1*3)+(x^2)/(2*3^2)+(x^3)/(3*3^3)+...+(x^n)/(n*3^n)+...的收敛域.2.将函数 f(x)=1/(x^2+3x+2)展开成x的幂级数.请看清题,电脑打字不太方便表达.只有这点财富值了,回答完整的再加分.
1、通项的系数an=1/(n*3^n),a(n+1)/an=n/(3n+3)→1/3(n→∞),所以收敛半径R=1/(1/3)=3,收敛区间是(-3,-3).x=3时,幂级数变为∑1/n,发散.x=-3时,幂级数变为∑1(-1)^n/n,由莱布尼兹定理,级数收敛.所以,收敛域是[-3,3).2、f(x)=1/((x+1)(x+2))=1/(1+x)-1/(2+x)=1/(1+x)-1/2×1/(1+x/2)1/(1+x)=∑(-1)^n*x^n,-1<x<1.1/(1+x/2)=∑(-1)^n*x^n/2^n,-2<x<2.所以,f(x)=∑(-1)^n*x^n - 1/2×∑(-1)^n*x^n/2^n=∑(-1)^n*(1-1/2^(n+1))x^n.收敛范围是-1<x<1.
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扫描下载二维码已知数列{an}满足a1=x,a2=3x,n+1+Sn+Sn-1=3n2+2(n≥2,n∈N*),Sn是数列{an}的前n项和.(ⅰ)求数列的通项an;(1)若数列{an}为等差数列.(ⅱ)若数列{bn}满足n=2an,数列{cn}满足n=t2bn+2-tbn+1-bn,试比较数列{bn}前n项和Bn与{cn}前n项和Cn的大小;(2)若对任意n∈N*,an<an+1恒成立,求实数x的取值范围.【考点】.【专题】等差数列与等比数列;不等式的解法及应用.【分析】(1)()由知可得,n+1+an+an-1=3n2+2-(3n2-6n+5)=6n-3结合等中项性即可求出数列的通项公式a;(ⅱ)根据ⅰ)可知n=2an=22n-1,n=t2bn+2-tbn+1-bn=16t2-4-1)bn.从而B=b12+…b,=c1+c2+…+cn=(1624t-1)(1+b2+…+n).只需比t4t-与1的大小可出n与Cn大关系;(2)利用已知件得an+an=6(n≥2,n∈N).然后分nk-1n=3,n3k+三情况讨论,列等式组答可.【解答】解:(1)(ⅰ)∵n+1+Sn+Sn-1=3n2+2(n≥2,n∈N*),①∴3an=6n-3.(2)∵n+1+Sn+Sn-1=3n2+2(n≥2,n∈N*),④∴当n=1时,an=a1=x.an=a3k=a3+(k-1)×6∴数列{an}的通项公式为an=2n-1.a1=1,a2=3符合③式.当16t2-4t-1>1,即t>或t<时,Bn<Cn.=1+6x+6k-6=2n-9x+8.①-②,得an=a3k+1=a4+(k-1)×6当n=3k+1时,an=a3k-1=a2+(k-1)×6=(16t2-4t-1)(b1+b2+…+bn).∴a1<a2且a3k-1<a3k<a3k+1<a3k+2.n+1+an+an-1=3n2+2-(3n2-6n+5)=6n-3.(ⅱ)∵an=2n-1.∴an+1+an-1=2an.=14-9x+6k-6∴Bn=b1+b2+…+bn,当n=3k时,an+3-an=6(n≥2,n∈N*).∴an=2n-1(n≥3)③解得,.∵对任意n∈N*,an<an+1恒成立,∴n=t2bn+2-tbn+1-bn∵数列{an}为等差数列,Cn=c1+c2+…+cn⑥-⑦,得=2n+3x-4.∴实数x的取值范围为.【点评】本考查差列,等比列的性,数列与式的综合问题的答等知识,属于题.声明:本试题解析著作权属菁优网所有,未经书面同意,不得复制发布。答题:zhtiwu老师 难度:0.46真题:2组卷:28
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